library(tidyverse)
library(openintro)
population <- ames$Gr.Liv.Area
samp <- sample(population, 60)
samp
##  [1] 1210 1764 1513 1567 1082 1341  864 1838 1112 1425 1211 1812  912 1134 1496
## [16] 1262 1253 1340 1355 2322 1694 1652 1760 2495 1876 1734 1226  848 1660 1208
## [31] 1537 2443 1830 2318 1621 1308 3238 2296 2263 1386 1594 1377 1025 1240 2076
## [46] 1713 1989 1280  788  860 1464 1258 2061 1856 1852 2128  951  616 1692  819
hist(samp)

mean(samp)
## [1] 1547.417

Exercise 1

Describe the distribution of your sample. What would you say is the “typical” size within your sample? Also state precisely what you interpreted “typical” to mean.

The distribution has a left skewing slope, centered around the mean of 1421 square feet for gross living area. There are small amounts of outliers approaching 6000 square feet on the right tail, and a hard cutoff of 660 for a minimum. The median gross living area is 1420, which is right near center in my sample.

Exercise 2

Would you expect another student’s distribution to be identical to yours? Would you expect it to be similar? Why or why not?

I would expect each sample of 60 to be slightly different distributions from mine. My sample was nearly exactly center of the mean, indicating that other students’ samples would be less and more than my mean, and their minimums and maximums may look different as well. The general shape, skew, approximate minimums and maximums would be expected to hold similar interpretations.

Exercise 3

For the confidence interval to be valid, the sample mean must be normally distributed and have standard error s/√n. What conditions must be met for this to be true?

The sample observations must be random, and independently chosen from one another. This is done randomly in the 60 designation of the Rstudio code. The size of the sample also must be less than 10% of the total population, ensuring more repeatability that future samples will also be differentiated from one another. The population was 2930, so a maximum sample size of 293 would allow for confidence intervals. My sample of 60 lies within this range to satisfy confidence intervals.

# Insert code for Exercise 3 here
sample_mean <-mean(samp)
qnorm(.975)
## [1] 1.959964
se <- sd(samp) / sqrt(60)
lower <- sample_mean - 1.96 * se
upper <- sample_mean + 1.96 * se
c(lower, upper)
## [1] 1420.704 1674.130

Exercise 4

What does “95% confidence” mean?

95 percent confidence interval means that upon repetitions of random sampling, we would expect 95% of the samples to lie between two values.

# Insert code for Exercise 4 here
mean(population)
## [1] 1499.69

Exercise 5

Does your confidence interval capture the true average size of houses in Ames?

My 95% confidence interval was 1306 - 1536 gross living area. The population mean was found to be 1499. 1499 of the population mean

# Insert code for Exercise 5 here
mean(population)
## [1] 1499.69

Exercise 6: EXTRA CREDIT

Each student in your class should have gotten a slightly different confidence interval. What proportion of those intervals would you expect to capture the true population mean? Why?

I would expect 95% of the responses to contain the true population mean of 1499. This would mean that 19 of 20 students would have confidence intervals containing the population mean. 1 in 20 students could be expected to have taken a sample that yields a confidence interval that does not include the true population mean.

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